Section 5.1 Mathematical Induction For each of the following...

Section 5.1 Mathematical Induction For each of the following exercises follow the given steps. (a) Determine and prove the Basis Step. (b) For the Inductive Step, write clearly the hypothesis and the thesis. (c) Prove the inductive step. Exercise 2. Prove by induction that, for all positive integers n, we have the following inequality: ?_{i=1}^{n} (-1)^i i^2 = [(-1)^n n(n + 1)] / 2 Exercise 3. Prove by induction that, for all integers n ? 2, we have ?_{i=1}^{n} 1/i^2 < 2 - 1/n Exercise 4. Prove by induction that ?_{i=1}^{n} i · (i!) = (n + 1)! - 1 whenever n ? Z^+. Exercise 5. Prove by induction that, for all positive integers n, we have n + 1 ? 2^n. Hint: you can use the fact that for all positive integers n, we have 1 ? 2^n. Note: you need to show where you use the hint.

Transcript

00:01 Hello everyone, here let's consider that pn is equal to summation i is equals to 1 to n minus 1 to the power i multiplied by i square so that is basically equals to minus 1 square plus 2 square minus 3 square plus 4 square and so on till minus 1 to the power n multiplied by n square so that is equals to minus 1 place to the power n multiplied by n plus 1 divided by 2 so let's say this is our equation number 1 so now for the basis of induction for the basis of induction let's say that the equation number 1 is true for n equals 1 so we'll have p of 1 will be equal to minus 1 to the power 1 that is equals to minus 1 and the right hand side would be equals to minus 1 to the power 1 multiplied by 1 multiplied by 1 plus 1 divided by 2 so from here we'll get this value is also equals to 1 so we can say that the p1 is true.

01:30 Now let's see the next part.

01:31 Now here for the second part for inductive hypothesis we have let's choose some k that belongs to n so let's assume that p k holds so we have p of k would be equals to minus 1 square plus 2 square minus 3 square plus 4 square and so on till minus 1 to the part k multiplied by k square and this is basically equals to minus 1 to the part k multiplied by k multiplied by k plus 1 divided by 2 so now here if we add minus 1 to the power k plus 1 multiplied by k plus 1 to the whole square to the both sides then we get minus 1 square plus 2 square minus 3 square minus and so on till plus minus 1 to the power k multiplied by k square plus minus 1 to the power k plus 1 multiplied by k plus 1 square and that is equals to minus 1 to the power k multiplied by k multiplied by k plus 1 divided by 2 plus minus 1 to the power k plus 1 multiplied by k plus 1 whole square so on simplifying it we will get this will be equal to k plus 1 multiplied by minus 1 to the power k multiplied by k by 2 minus k plus 1 and that would be equals to k plus 1 multiplied by minus 1 to the power k multiplied by k minus 2k minus 2 divided by 2 so further more simplifying it we'll get this would be equals to k plus 1 multiplied by minus 1 to the power k minus k k minus 2 divided by 2 and that would be equals to k plus 1 multiplied by minus 1 to the power k multiplied by minus 1 multiplied by k plus 2 divided by 2 so further more simplifying it we'll get this would be equals to minus 1 to the power k multiplied by k plus 1 multiplied by k plus 2 divided by 2 so here here we have got that p of k plus 1 is also true...

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